# Complex function defined as an integral: is it analytic?

I'm trying to study for my final this Thursday and for some reason the problem that is giving me the most trouble is, according to the professor, the "easiest one". I'm just not seeing something here, it's been a long time since I've done anything in complex that looks like this. The problem is:

$$A = \left\{ \zeta: |\zeta|= \frac{1}{2}\right\}, B: = \left\{\omega : |\omega|< \frac{3}{4}\right\}. \text{ Define } f(w) : = \oint_A \frac{z(z+1)}{z^2 + 2z -w} dz.$$

I'm asked to show that $f$ is analytic in $B$ and to find $f'(0)$. I need to learn the idea behind this much more than I need an answer to this specific question, so hints or related questions are much appreciated.

Thank you fixed!

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It seems like there may be something missing in the problem. First, you haven't said what the contour you're integrating over is. Second, I don't see where the set $A$ comes into the problem. The question never makes any mention of $A$. –  froggie Dec 12 '12 at 2:02
Are there any poles in the interior of $A$? –  alex.jordan Dec 12 '12 at 3:19
This seems like a hard question to answer, as it depends on what $\omega$ we are evaluating at. The for example, if $\omega = \frac{2}{3}$ then there is a pole at $z = \frac{1}{2} ( \sqrt{7} - 2)$. But of course there will be different poles for different $\omega$. Are you suggesting I do something with the residues as a function of $\omega$? –  abstruse Dec 12 '12 at 3:33

## 2 Answers

The value of the integral will sum the residues at the poles that are interior to $A$. For $\omega$ in the restricted region, the denominator has two distinct roots, so there are either $0,1$, or $2$ simple poles within $A$ and its interior.

The roots of $z^2+2z-\omega$ are $-1\pm\sqrt{1+\omega}$. Let $U=\{\omega: |\omega|<3/4\}$. Then $\sqrt{1+U}$ is two disjoint regions in $\mathbb{C}$, one with positive real part and one with negative real part. So of the two roots of $z^2+2z-\omega$, one is to the left of $-1$ and exterior to $A$.

Certainly, there are values of $\omega$ for which the integrand has a pole at $c=-1+\sqrt{1+\omega}$ within $A$. The residue at a simple pole $c$ is \begin{aligned} \lim_{z\to c}(z-c)\frac{z(z+1)}{z^2+2z-\omega}&=\lim_{z\to c}(z-c)\frac{z(z+1)}{(z-c)(z+\omega/c)}\\ &=\frac{c(c+1)}{c+\omega/c}\\ &=\frac{c(c^2+c)}{c^2+\omega}\\ &=\frac{c(\omega-c)}{2(\omega-c)}\\ &=\frac{(-1+\sqrt{1+\omega})(\omega+1-\sqrt{1+\omega})}{2(\omega+1-\sqrt{1+\omega})}\\ \end{aligned}

Define $g(\omega)$ to be this last expression.

So $f(w)=g(w)$ if $\omega$ provides one pole within $A$. If the claim is to be true, then $f(w)=g(w)$ for all $w$ in $U$. The only way that the claim could fail to be true is if for some $\omega_0\in U$, the integrand has no poles within $A$ (making $f(\omega_0)=0$). (Although this would not be a problem if $\lim{\omega\to\omega_0}g(\omega)$ equals $0$ too.)

As noted in the comments, at $\omega=0$ there are no interior poles, but that's OK since the $g(\omega)$ approaches $0$ as $\omega\to0$.

It remains to show that when $\omega\in U\setminus\{0\}$, then $|-1+\sqrt{1+\omega}|<1/2$, and therefore there will be one pole interior to $A$. Let $V=\{c: |c|<1/2\}$. We are trying to show that $k(U)\subset V$, where $k(\omega)=-1+\sqrt{1+\omega}$. Since $k$ is one-to-one on $U$ and $V$, then this is equivalent to showing $U\subset k^{-1}(V)=\{c^2+2c:|c|<1/2\}$. My argument is that the boundary curve for $U$ ($z=\frac{3}{4}e^{it})$ and the boundary curve for $k^{-1}(V)$ ($z=\frac{1}{4}e^{2it}+e^{it}$) intersect tangentially at $t=\pi$ (corresponding to $z=-3/4$) and do not intersect anywhere else. Comparing $z$-values where $t=0$ reveals that $U$'s boundary is the interior curve.

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Is this true? It seems like at $\omega=0$ there are two poles, one at $z=0$ and one at $z=−2$. Thus there is still exactly one pole inside of A. Am I missing something here? –  abstruse Dec 12 '12 at 6:09
@alexjordan: why do you say that it is just the number of poles inside this disk times $2\pi i$ and not something to do with the sum of the residues inside $A$? –  user52828 Dec 12 '12 at 17:32
@abstruse If $\omega=0$,then the integrand is $\frac{z+1}{z+2}$ after cancellation. –  alex.jordan Dec 14 '12 at 7:35
@user52828 You're right - I was oversimplifying. Let me see if I can adjust. –  alex.jordan Dec 14 '12 at 7:38

Let's write $$f(w)=\oint_A \frac{g(z)}{h(z)-w}\,dz,\quad g(z)=z^2+z, \quad h(z)=z^2+2z$$ When $|w|<3/4$, we have $|2z|>|z^2-w|$ on $A$. By Rouche's theorem there is a unique root of $h(z)=w$ within the region $D$ bounded by $A$. Write it as $h^{-1}(w)$. Since $h'(z)=2z+2$ does not vanish in $D$, it follows that $h^{-1}(w)$ is a holomorphic function of $w\in B$.

By the residue theorem, $$f(w)=2\pi i \frac{g'(h^{-1}(w))}{h'(h^{-1}(w))}\tag1$$ which is evidently a holomorphic function in $B$. Before calculating $f'$ it is helpful to plug in explicit $g,h$ and simplify: $$f(w)=2\pi i\frac{2h^{-1}(w)+1}{2h^{-1}(w)+2}=2\pi i -\frac{\pi i}{h^{-1}(w)+1} \tag2$$ Now take the derivative using the chain rule and the inverse function theorem: $$f'(w) = \frac{\pi i}{(h^{-1}(w)+1)^2} \frac{d(h^{-1}(w))}{dw} =\frac{\pi i}{(h^{-1}(w)+1)^2 h'(h^{-1}(w))} = \frac{\pi i/2}{(h^{-1}(w)+1)^3 } \tag3$$ Finally, plug in $w=0$, which yields $h^{-1}(0)=0$ and $f'(0) = \pi i/2$.

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